\documentclass[ 11pt, reqno ]{amsart} \usepackage{mathtools} \usepackage{amssymb} \usepackage{amsthm} \usepackage{mathrsfs} \usepackage{microtype} \usepackage[ backend=biber, style=numeric, doi=true, sorting=nyt, eprint=true, giveninits=true, maxnames=99 ]{biblatex} \addbibresource{bib/references.bib} \DeclareFieldFormat[article]{title}{\mkbibemph{#1}} \renewbibmacro*{in:}{% \ifentrytype{article}{}{\printtext{\bibstring{in}\intitlepunct}}} \usepackage[ colorlinks=false, linkbordercolor={1 0 0}, citebordercolor={0 1 0}, urlbordercolor={0 1 1} ]{hyperref} \usepackage[nameinlink,noabbrev]{cleveref} \newcommand{\Exp}{\operatorname{Exp}} \newcommand{\dH}{d_{\mathrm H}} \newcommand{\dGH}{d_{\mathrm{GH}}} \newcommand{\uGH}{u_{\mathrm{GH}}} \newcommand{\diam}{\operatorname{diam}} \newcommand{\dis}{\operatorname{dis}} \newcommand{\cPp}{\mathcal{P}_{+}} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \theoremstyle{plain} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{question}[theorem]{Question} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \crefname{theorem}{theorem}{theorems} \crefname{lemma}{lemma}{lemmas} \crefname{proposition}{proposition}{propositions} \crefname{corollary}{corollary}{corollaries} \crefname{definition}{definition}{definitions} \crefname{example}{example}{examples} \crefname{question}{question}{questions} \crefname{remark}{remark}{remarks} \Crefname{theorem}{Theorem}{Theorems} \Crefname{lemma}{Lemma}{Lemmas} \Crefname{proposition}{Proposition}{Propositions} \Crefname{corollary}{Corollary}{Corollaries} \Crefname{definition}{Definition}{Definitions} \Crefname{example}{Example}{Examples} \Crefname{question}{Question}{Questions} \Crefname{remark}{Remark}{Remarks} \newenvironment{acknowledgements}{% \medskip\noindent\textit{Acknowledgements.}\ }{\par} \newenvironment{useofai}{% \medskip\noindent\textbf{Use of AI.}\ }{\par} \title[Finite Hausdorff Hyperspaces] {Finite Hausdorff Hyperspaces and Gromov--Hausdorff Geometry} \author{Yoshito Ishiki} \date{30 September 2026} \subjclass[2020]{Primary 54B20; Secondary 51F30, 54E35} \keywords{Hausdorff hyperspace, Gromov--Hausdorff distance, Gromov--Hausdorff ultrametric, strongly rigid metric space, ultrametric space} \begin{document} \begin{abstract} For a nonempty finite metric space $X$, let $\Exp(X)$ be the space of its nonempty subsets equipped with the Hausdorff metric. We prove that the isometry type of $\Exp(X)$ determines $X$ when $X$ is strongly rigid, meaning that distinct unordered pairs of distinct points have distinct distances. We also prove that the hyperspace operation preserves the ordinary Gromov--Hausdorff distance locally under explicit separation conditions on the distance spectra. For finite ultrametric spaces, the hyperspace operation is an isometric embedding with respect to the Gromov--Hausdorff ultrametric. In contrast, we construct finite ultrametric spaces for which the ordinary Gromov--Hausdorff distance strictly decreases under the hyperspace operation. Examples of equal cardinality exist in every cardinality at least six and realize every ratio in $[1/2,1)$. Their distances remain constant after the first hyperspace iteration. An explicit perturbation also yields strict contraction for finite strongly rigid metric spaces with all triangle inequalities strict. \end{abstract} \maketitle \tableofcontents \section{Introduction} Let $\Exp(X)$ denote the Hausdorff hyperspace of a compact metric space $X$, that is, the space of its nonempty compact subsets with the Hausdorff metric. Mikhailov proved that the assignment \[ X\longmapsto \Exp(X) \] is nonexpanding on the Gromov--Hausdorff space \cite{Mikhailov2018}. It is therefore natural to ask whether the assignment preserves every Gromov--Hausdorff distance. We first study whether the hyperspace isometry type determines the base. A finite metric space is strongly rigid if distinct unordered pairs of distinct points have distinct distances. We prove that if $X$ and $Y$ are finite strongly rigid metric spaces, then \[ \Exp(X)\cong\Exp(Y) \quad\Longrightarrow\quad X\cong Y. \] One proof recognizes singleton subsets from the cardinalities of their closed balls in the hyperspace. For a strongly rigid base with $n$ points, a second proof reconstructs the base from any family of $n$ hyperspace points whose $\binom n2$ distances are all different. These arguments also imply quantitative local results for the ordinary distance. In a symmetric form, $\Exp$ preserves $\dGH(X,Y)$ when it is smaller than one quarter of the minimum gap in the two finite distance spectra. For bases with equal cardinalities, the gap condition can be imposed on only one of the two spaces. Thus the ordinary hyperspace operation has explicit conditions for local distance preservation even though it is not globally isometric. The answer depends on which Gromov--Hausdorff geometry is used. On finite ultrametric spaces we consider the Gromov--Hausdorff ultrametric $\uGH$, introduced by Zarichnyi \cite{Zarichnyi2005}. For all nonempty finite ultrametric spaces $U,V$, we prove that \[ \uGH(\Exp(U),\Exp(V))=\uGH(U,V). \] Thus $\Exp$ is an isometric embedding in the non-Archimedean setting. The proof has two ingredients. First, a finite ultrametric space can be reconstructed from the isometry type of its Hausdorff hyperspace. Second, taking a Hausdorff hyperspace commutes with every closed quotient of a finite ultrametric space. The structural formula for $\uGH$ due to M\'{e}moli, Smith, and Wan \cite{MemoliSmithWan2021} then yields the distance equality. Here ``non-Archimedean'' refers to the distance $\uGH$, not merely to restricting the ordinary Gromov--Hausdorff distance $\dGH$ to ultrametric spaces. This distinction is essential. We construct finite ultrametric spaces $X_0,Y_0$ such that \[ 2\dGH(X_0,Y_0)=3, \qquad 2\dGH(\Exp(X_0),\Exp(Y_0))=2. \] The contraction follows from the sizes of the components at the largest distance. The three top components of $X_0$ all have two points, whereas those of $Y_0$ have sizes $1,1,3$. At the base level a two-point component cannot cover a three-point component below distortion three. After passing to hyperspaces, the seven support blocks have sizes \[ (3,3,3,9,9,9,27) \quad\text{and}\quad (1,1,1,7,7,7,7), \] and the former blocks cover the latter at distortion two. The same phenomenon persists at equal cardinality. For every $00 \quad\Longrightarrow\quad \{x,x_1\}=\{y,y_1\}. \] Thus all unordered pairs of distinct points determine different distances. Strong rigidity first yields an exact reconstruction result and then, after a quantitative separation argument, local preservation of the ordinary Gromov--Hausdorff distance. For a nonempty finite metric space $M$, put \[ \Sigma(M)=\{d_M(x,x_1)\mid x,x_1\in M\}. \] \begin{lemma} \label{lem:finite-hyperspace-spectrum} Let $M$ be a nonempty finite metric space. Then \[ \Sigma(\Exp(M))=\Sigma(M). \] \end{lemma} \begin{proof} Take $s\in\Sigma(M)$ and choose $x,y\in M$ with $s=d_M(x,y)$. Since $\dH^M(\{x\},\{y\})=s$, we have $s\in\Sigma(\Exp(M))$. Conversely, take $s\in\Sigma(\Exp(M))$ and choose $A,B\in\Exp(M)$ with $s=\dH^M(A,B)$. The defining maxima and minima are attained because $A$ and $B$ are finite. Thus $s=d_M(a,b)$ for some $a\in A$ and $b\in B$, and hence $s\in\Sigma(M)$. \end{proof} \subsection{Exact reconstruction} We recognize singleton subsets by counting balls in the hyperspace. Let $X$ be a finite metric space. For $A\in\Exp(X)$ and $r\ge0$, define the closed-ball profile \[ \beta_A^X(r) = \#\{B\in\Exp(X)\mid\dH^X(A,B)\le r\}. \] This profile recognizes the singleton locus intrinsically. \begin{theorem} \label{thm:singleton-characterization} Let $X$ be a finite strongly rigid metric space with $\lvert X\rvert\ge3$, and let $A\in\Exp(X)$. Then $A$ is a singleton if and only if $\beta_A^X(r)+1$ is a power of two for every $r\ge0$. \end{theorem} \begin{proof} For a nonempty set $S\subseteq X$ and $r\ge0$, write \[ N_r(S)=\{x\in X\mid d_X(x,S)\le r\}. \] If $A=\{a\}$, then for every $B\in\Exp(X)$ we have \[ \dH^X(\{a\},B)=\max_{b\in B}d_X(a,b). \] Consequently, the sets $B$ counted by $\beta_{\{a\}}^X(r)$ are precisely the nonempty subsets of $N_r(\{a\})$, and hence \[ \beta_{\{a\}}^X(r)+1=2^{\lvert N_r(\{a\})\rvert}. \] Conversely, assume that $\lvert A\rvert\ge2$. Among the positive distances having at least one endpoint in $A$, let $R$ be the largest. Strong rigidity yields a unique pair $\{u,v\}$ realizing $R$. There is at least one further distance having an endpoint in $A$, because $\lvert X\rvert\ge3$. Let $r0. \] By \cref{lem:finite-hyperspace-spectrum}, this is also the minimum gap between distinct distance levels of $\Exp(M)$. \begin{theorem} \label{thm:local-strongly-rigid-isometry} Let $X,Y$ be finite strongly rigid metric spaces with $\lvert X\rvert,\lvert Y\rvert\ge3$, and put $\eta=\min\{\eta(X),\eta(Y)\}$. If \[ \dGH(X,Y)<\frac{\eta}{4}, \] then \[ \dGH(\Exp(X),\Exp(Y))=\dGH(X,Y). \] \end{theorem} \begin{proof} Put \[ g=\dGH(X,Y), \qquad h=\dGH(\Exp(X),\Exp(Y)). \] By \cref{prop:ordinary-nonexpansion}, $h\le g<\eta/4$. Since the hyperspaces are finite, there is an optimal correspondence $R\subseteq\Exp(X)\times\Exp(Y)$ with \[ \delta=\dis(R)=2h<\frac{\eta}{2}. \] If one source set were related to two distinct target sets, their positive Hausdorff distance would be at most $\delta<\eta(Y)$, contradicting the definition of $\eta(Y)$. The symmetric argument rules out two distinct source sets related to one target set. Thus $R$ is the graph of a bijection $F:\Exp(X)\to\Exp(Y)$. Fix a singleton $A\in\Exp(X)$, and let $r\in\Sigma(Y)$ be a nonmaximal distance level with successor $r^+$. For each $C\in\Exp(X)$, if $\dH^Y(F(A),F(C))>r$, then \[ \dH^Y(F(A),F(C))\ge r^+\ge r+\eta, \] and the distortion bound yields \[ \dH^X(A,C)\ge r+\eta-\delta>r+\delta. \] Together with the direct distortion implication, this shows that $F$ satisfies \[ \begin{aligned} &F\bigl(\{C\in\Exp(X)\mid\dH^X(A,C)\le r+\delta\}\bigr)\\ &\qquad=\{D\in\Exp(Y)\mid\dH^Y(F(A),D)\le r\}. \end{aligned} \] The same equality of cardinalities is automatic at the maximal level. Closed-ball counts are constant between consecutive spectrum levels, so the profile of $F(A)$, plus one, is a power of two at every radius. Hence \cref{thm:singleton-characterization} shows that $F(A)$ is a singleton. Applying the same argument to $F^{-1}$ proves that $F$ bijects the two singleton loci. The restriction of $R$ to these loci is therefore a base correspondence of distortion at most $\delta$. By \eqref{eq:correspondence-formula}, $g\le\delta/2=h$. The reverse inequality is nonexpansion. \end{proof} For fixed cardinality the separation can be imposed on only one of the two spaces. The proof uses a second reconstruction mechanism. \begin{theorem} \label{thm:rainbow-basis} Let $Y$ be a finite strongly rigid metric space with $n$ points. If an $n$-point subspace \[ \mathcal A=\{A_0,\ldots,A_{n-1}\}\subseteq\Exp(Y) \] has pairwise distinct distances on its $\binom n2$ unordered pairs, then $\mathcal A\cong Y$. \end{theorem} \begin{proof} The case $n=1$ is immediate, so assume $n\ge2$. By \cref{lem:finite-hyperspace-spectrum}, the $\binom n2$ distinct positive distances in $\mathcal A$ exhaust the positive distance spectrum of $Y$. Let $D=\diam(Y)$. A pair in $\mathcal A$ is at Hausdorff distance $D$. This distance is realized by a directed nearest-point term, so after interchanging the two sets there are $p\in A$ and $B\in\mathcal A$ such that $d_Y(p,B)=D$. Every point of $B$ is then at distance $D$ from $p$. Strong rigidity yields a unique diameter pair $\{p,q\}$, whence $B=\{q\}$. Relabel so that $A_0=\{q\}$. For $i>0$, put \[ r_i=\dH^Y(\{q\},A_i)=\max_{a\in A_i}d_Y(q,a). \] These $n-1$ distinct values exhaust the distances from $q$. Relabel the remaining sets and write \[ Y=\{q,y_1,\ldots,y_{n-1}\}, \qquad r_1<\cdots1$ be the least degree of a remaining factor. For every $m\in\mathscr M$ of degree $d$, the coefficient of $m$ in the product is exactly the multiplicity with which $m$ occurs as a factor. A product of two or more remaining factors has degree at least $d^2>d$. Hence the two sides have the same multiset of degree- $d$ factors. These common factors may be cancelled even when $\mathscr M$ is not a cancellative monoid. To see this, let $F\in\mathbb N[\mathscr M]$ have identity coefficient one and all other monomials of degree greater than one. Let $G,H\in\mathbb N[\mathscr M]$ satisfy $FG=FH$. For the sake of contradiction, suppose that $G\ne H$. Pass to $\mathbb Z[\mathscr M]$ and choose the least degree $n$ at which $G-H$ has a nonzero coefficient. Choose a monomial $m$ of degree $n$ whose coefficient in $G-H$ is nonzero. Then every contribution to the coefficient of $m$ coming from a nonidentity monomial of $F$ uses a coefficient of $G-H$ of degree strictly smaller than $n$. Those coefficients vanish, so the coefficient of $m$ in $F(G-H)$ equals its nonzero coefficient in $G-H$, a contradiction. Thus multiplication by $F$ is injective. Cancel all degree- $d$ factors and repeat. Finiteness terminates the procedure and recovers the complete factor multiset. \end{proof} \begin{theorem} \label{thm:finite-ultrametric-reconstruction} Let $U,V$ be nonempty finite ultrametric spaces. If \[ \Exp(U)\cong\Exp(V), \] then $U\cong V$. \end{theorem} \begin{proof} We argue by induction on the common cardinality of $U$ and $V$, which is determined by $\lvert\Exp(U)\rvert=2^{\lvert U\rvert}-1$. The one-point case is immediate. Otherwise, \[ \diam(U)=\diam(\Exp(U))=\diam(\Exp(V))=\diam(V)=D>0. \] Write the open $D$-partitions as \[ U=U_1\sqcup\cdots\sqcup U_k, \qquad V=V_1\sqcup\cdots\sqcup V_\ell. \] An isometry of the hyperspaces preserves their open $D$-blocks and the isometry type of each block. By \eqref{eq:support-max-product}, equality of these block multisets, after adjoining the empty-support term, is \begin{equation} \prod_{i=1}^k\left(1+[\Exp(U_i)]\right) = \prod_{j=1}^\ell\left(1+[\Exp(V_j)]\right). \label{eq:hyperspace-factorization} \end{equation} In \eqref{eq:hyperspace-factorization}, brackets denote isometry classes in the commutative monoid of nonempty finite ultrametric spaces under $\boxtimes$. Equip this monoid with the degree $\deg[W]=\lvert W\rvert$. Its identity is the one-point space, so the hypotheses of \cref{lem:graded-subset-product-inversion} hold. The lemma applied to \eqref{eq:hyperspace-factorization} matches the factors $\Exp(U_i)$ and $\Exp(V_j)$ up to isometry. Every top block has smaller cardinality than its ambient space. The induction hypothesis therefore yields a matching isometry $U_i\cong V_j$ for every factor. Their union is an isometry $U\cong V$, since distances between distinct top blocks equal $D$. \end{proof} We next compare the closed quotients of $U$ and $\Exp(U)$. For a finite ultrametric space $U$ and $t\ge0$, let $q_t:U\to U_{\mathfrak c(t)}$ be the quotient map. \begin{proposition} \label{prop:closed-quotient-commutation} Let $U$ be a finite ultrametric space and let $t\ge0$. Then the map \[ Q_t: (\Exp(U))_{\mathfrak c(t)} \to \Exp(U_{\mathfrak c(t)}), \qquad [A]\longmapsto q_t(A), \] is a well-defined isometry. \end{proposition} \begin{proof} For nonempty subsets $A,B\subseteq U$, we have \begin{equation} \dH^U(A,B)\le t \quad\Longleftrightarrow\quad q_t(A)=q_t(B). \label{eq:quotient-support-equivalence} \end{equation} The Hausdorff bound says exactly that each closed $t$-class met by one set is met by the other. Thus $Q_t$ is well-defined and injective. It is surjective because representatives may be chosen from the members of any nonempty subset of the finite quotient. If $q_t(A)\ne q_t(B)$, at least one quotient class is met on only one side. Its Hausdorff contribution is greater than $t$, while the contribution of a class met on both sides is at most $t$. Distances between distinct classes are their quotient distances. Taking the maximum yields \[ \dH^U(A,B) = \dH^{U_{\mathfrak c(t)}}(q_t(A),q_t(B)). \] Together with \eqref{eq:quotient-support-equivalence}, this proves that $Q_t$ is an isometry. \end{proof} \begin{theorem} \label{thm:finite-ugh-isometry} Let $U,V$ be finite ultrametric spaces. Then \[ \uGH(\Exp(U),\Exp(V))=\uGH(U,V). \] Consequently, $\Exp$ is an isometric embedding on the isometry classes of nonempty finite ultrametric spaces equipped with $\uGH$. \end{theorem} \begin{proof} The structural formula \eqref{eq:ugh-structural}, \cref{prop:closed-quotient-commutation}, and \cref{thm:finite-ultrametric-reconstruction} yield \begin{align*} \uGH(\Exp(U),\Exp(V)) &= \min\left\{ t\ge0\,\middle|(\Exp(U))_{\mathfrak c(t)} \cong(\Exp(V))_{\mathfrak c(t)} \right\} \\ &= \min\left\{ t\ge0\,\middle|\Exp(U_{\mathfrak c(t)}) \cong\Exp(V_{\mathfrak c(t)}) \right\} \\ &= \min\left\{ t\ge0\,\middle|U_{\mathfrak c(t)}\cong V_{\mathfrak c(t)} \right\} \\ &= \uGH(U,V). \end{align*} The reconstruction theorem also shows directly that the map on isometry classes is injective. \end{proof} \begin{remark} The theorem concerns the Gromov--Hausdorff ultrametric $\uGH$. It makes no claim that the ordinary distance $\dGH$ is preserved on ultrametric bases. The next section shows that this stronger-looking interpretation is false. \end{remark} \section{Strict contraction for the ordinary distance} \label{sec:ordinary-contraction} We now show that \cref{thm:finite-ugh-isometry} has no analogue for the ordinary Gromov--Hausdorff distance. The first example is ultrametric and has only two positive distance levels in each space. \subsection{An exact ultrametric counterexample} Let \[ X_0=P_1\sqcup P_2\sqcup P_3, \qquad \lvert P_1\rvert=\lvert P_2\rvert=\lvert P_3\rvert=2. \] Equip $X_0$ with the ultrametric whose distance is $2$ between distinct points of the same $P_i$ and $10$ between points in different components. Let \[ Y_0=Q_1\sqcup Q_2\sqcup Q_3, \qquad (\lvert Q_1\rvert,\lvert Q_2\rvert,\lvert Q_3\rvert)=(1,1,3). \] The three distinct points of $Q_3$ are at distance $3$, and points in different $Q_j$ 's are at distance $10$. \begin{theorem} \label{thm:ordinary-ultrametric-contraction} The spaces $X_0,Y_0$ defined above satisfy \[ 2\dGH(X_0,Y_0)=3, \qquad 2\dGH(\Exp(X_0),\Exp(Y_0))=2. \] In particular, \[ \dGH(\Exp(X_0),\Exp(Y_0)) < \dGH(X_0,Y_0). \] \end{theorem} \begin{proof} We first compute the base distance. Collapse $P_1$ and $P_2$ onto the singleton components $Q_1$ and $Q_2$. If $P_3=\{a,b\}$ and $Q_3=\{u,v,w\}$, use the component relation \[ \{(a,u),(a,v),(b,w)\}. \] The first two component relations have distortion $2$. The third has distortion $3$, attained by the two relation edges with source $a$, and comparisons between different components compare $10$ with $10$. Thus there is a base correspondence of distortion $3$. For the sake of contradiction, suppose that a base correspondence $R$ had distortion strictly less than $3$. Any two targets related to source points in one $P_i$ would have distance less than \[ 2+3=5<10. \] They must therefore lie in one target component $Q_j$. Coverage of $Y_0$ induces a surjection from the three source components to the three target components, hence a bijection. One two-point component $P_i$ must then cover all three points of $Q_3$. Some source point is related to two distinct points of $Q_3$, producing distortion $3$, a contradiction. By \eqref{eq:correspondence-formula}, this proves $2\dGH(X_0,Y_0)=3$. It remains to compute the hyperspace distance. For every nonempty $S\subseteq\{1,2,3\}$, let $E_S^X$ be the family of subsets of $X_0$ whose set of met components is exactly $S$. Each $\Exp(P_i)$ is a three-point equilateral space of side $2$, and \[ E_S^X \cong \mathop{\boxtimes}_{i\in S}\Exp(P_i). \] Thus $E_S^X$ is equilateral of side $2$, and the seven source-block cardinalities are \begin{equation} (3,3,3,9,9,9,27). \label{eq:source-support-sizes} \end{equation} On the target side, the three support blocks not meeting $Q_3$ are singletons. Each of the other four blocks is isometric to the seven-point equilateral space $\Exp(Q_3)$ of side $3$. The target-block cardinalities are therefore \begin{equation} (1,1,1,7,7,7,7). \label{eq:target-support-sizes} \end{equation} Pair the three size- $3$ source blocks with the three singleton target blocks. Pair the remaining source blocks, of sizes $9,9,9,27$, with the four size- $7$ target blocks. In each pair take the graph of a surjection from the source block onto the target block. A size- $3$ block collapsed onto a singleton has distortion $2$. For the other pairs, two distinct source points with the same image cost $2$, while two source points with distinct images cost $\abs{2-3}=1$. Every block relation consequently has distortion at most $2$. Different support blocks are at Hausdorff distance $10$ on both sides. The union of the seven block relations is therefore a correspondence $\mathcal Q$ with $\dis(\mathcal Q)\le2$. For the reverse bound, the source and target hyperspaces have cardinalities $63$ and $31$, respectively, and the minimum positive distance in $\Exp(X_0)$ is $2$. Given any hyperspace correspondence, choose one target partner for each of the $63$ source points. Two distinct source points share a target partner, so those two relation edges have distortion at least $2$. Hence every hyperspace correspondence has distortion at least $2$, completing the proof. \end{proof} \begin{remark} This example emphasizes the distinction between the two ambient geometries. The closed quotients of $X_0$ and $Y_0$ first become isometric at $t=3$, so $\uGH(X_0,Y_0)=3$. By \cref{thm:finite-ugh-isometry}, \[ \uGH(\Exp(X_0),\Exp(Y_0))=3 \] even though their ordinary Gromov--Hausdorff distance drops from $3/2$ to $1$. \end{remark} \subsection{Scale-quotient capacity} The previous proof compares the cardinalities of entire support blocks. For multiscale targets, the exact invariant at a prescribed distortion is a quotient cardinality. If $Z$ is a finite ultrametric space and $a\ge0$, write \[ c_a(Z)=\#\bigl(Z/(d\le a)\bigr). \] The relation $d\le a$ is an equivalence relation by the ultrametric inequality. \begin{lemma} \label{lem:scale-quotient-capacity} Let $P$ be a positive integer and let $E_P(a)$ be the $P$-point equilateral space of side $a>0$, and let $Z$ be a nonempty finite ultrametric space with $\diam(Z)\le2a$. Then there is a correspondence of distortion at most $a$ between $E_P(a)$ and $Z$ if and only if \[ P\ge c_a(Z). \] \end{lemma} \begin{proof} In a correspondence of distortion at most $a$, all target partners of one source point have mutual distance at most $a$. They therefore lie in one closed $a$-class. Covering all classes requires $P\ge c_a(Z)$. Conversely, map the $P$ source points onto the closed $a$-classes of $Z$, and relate each source point to every point in its assigned class. Target distances within a class are at most $a$, while target distances between distinct classes belong to $(a,2a]$. Comparison with source distances $0$ and $a$ yields distortion at most $a$. \end{proof} We apply the lemma support by support. Fix $a>0$ and a positive integer $k$. For each $1\le i\le k$, let $p_i$ be a positive integer. Let $X$ have $k$ top components \[ X_i=E_{p_i}(a), \] at common mutual distance $H$. Let $Y$ have $k$ nonempty finite-ultrametric top components $Y_j$, also at mutual distance $H$, and assume \[ \max_{1\le j\le k}\diam(Y_j)\le2a, \qquad H>2a. \] For nonempty supports $S,T\subseteq\{1,\ldots,k\}$, put \[ P_S=\prod_{i\in S}(2^{p_i}-1), \qquad Q_T(a)=\prod_{j\in T}(2^{c_a(Y_j)}-1). \] \begin{theorem} \label{thm:scale-support-criterion} Let $X,Y,a,H,P_S,Q_T(a)$ be as defined above. Then there is a correspondence of distortion at most $a$ between $\Exp(X)$ and $\Exp(Y)$ if and only if the nonempty supports admit a bijection $\phi$ such that, for every nonempty $S\subseteq\{1,\ldots,k\}$, \[ P_S\ge Q_{\phi(S)}(a). \] \end{theorem} \begin{proof} The source support block indexed by $S$ is the $P_S$-point equilateral space of side $a$. The target support block indexed by $T$ is the maximum product of the spaces $\Exp(Y_j)$, $j\in T$. It has diameter at most $2a$. For a finite ultrametric space $Z$ and a nonempty subset $A\subseteq Z$, its closed $a$-class in $\Exp(Z)$ is determined by the collection of closed $a$-classes of $Z$ met by $A$. Hence \[ c_a(\Exp(Z))=2^{c_a(Z)}-1. \] Closed quotient cardinalities multiply under maximum products, so the target support block has $Q_T(a)$ closed $a$-classes. A source support block cannot meet two target support blocks under a distortion- $a$ correspondence. Its internal diameter is at most $a$, whereas distinct target supports are at distance $H>2a$. Coverage induces a surjection on the two sets of $2^k-1$ nonempty supports, hence a bijection. Restriction to a paired block and \cref{lem:scale-quotient-capacity} yield the required inequalities. Conversely, use \cref{lem:scale-quotient-capacity} on each paired pair of support blocks and take the union of the resulting correspondences. Distances between distinct support blocks are $H$ on both sides. \end{proof} \subsection{Equal-cardinality counterexamples} The quotient criterion permits low-scale target fibers without charging their full cardinality. This removes the cardinality imbalance in the first counterexample. \begin{theorem} \label{thm:equal-cardinality-contraction} Let $s,r$ be integers with $2\le s\le r$, put $n=r+s-2$, and assume \[ 00. \] The remaining source blocks mapped to supports meeting the third target component contain a factor $A_n$ and at least one factor $A_s$ or $A_r$. Since $s,r\ge2$, their capacities are at least $A_{n+1}$. \Cref{thm:scale-support-criterion} yields hyperspace distortion at most $a$. For the reverse inequality, the two hyperspaces have the same cardinality and minimum positive distance $a$. A correspondence of distortion below $a$ would therefore be a bijection. It would preserve the seven $H$-support blocks and their cardinalities. Equality of the two support-block cardinality multisets would, by \cref{lem:graded-subset-product-inversion}, force equality of the top-component cardinality multisets. This is impossible because the largest target component has size $n+1$, whereas every source component has size at most $n$. \end{proof} The minimal choice $s=r=2$ yields two six-point spaces. \[ X=E_2(a)\sqcup E_2(a)\sqcup E_2(a), \qquad Y=E_2(a)\sqcup E_1(a)\sqcup E_3(b). \] The components in these disjoint unions are at mutual distance $H$. Appending the same number of singleton top components to both spaces preserves the proof. Hence every cardinality at least six admits examples with equal cardinalities, and varying $a/b$ realizes every ratio in $[1/2,1)$. A strongly rigid example with strict triangle inequalities is constructed in the last subsection. \subsection{Iteration and the half-ratio boundary} The spaces in \cref{thm:equal-cardinality-contraction} retain their first hyperspace distance under further iteration. \begin{proposition} \label{prop:iterated-counterexample-stability} Let $X,Y$ be the spaces in \cref{thm:equal-cardinality-contraction}. For each integer $m\ge0$, put \[ D_m= 2\dGH\bigl(\Exp^m(X),\Exp^m(Y)\bigr). \] Then \[ D_0=b, \qquad D_m=a \quad(m\ge1). \] \end{proposition} \begin{proof} The upper bound $D_m\le a$ follows from the first-iterate construction and nonexpansion. The two iterated spaces have equal cardinality and minimum positive distance $a$. Thus a correspondence of distortion below $a$ would be a bijection preserving their top support blocks. For a vector $\mathbf v=(v_1,\ldots,v_k)$ of top-component cardinalities, the top-component cardinality multiset of its hyperspace is \[ F(\mathbf v)= \left\{ \prod_{i\in S}(2^{v_i}-1)\,\middle| \emptyset\ne S\subseteq\{1,\ldots,k\} \right\}. \] The graded subset-product inversion lemma says that $F$ is injective. The initial top-size vectors of $X$ and $Y$ differ, so their $F^m$-images differ for every $m$. A block-preserving bijection is therefore impossible, proving $D_m\ge a$. \end{proof} Finally, the threshold $1/2$ has a local explanation. Assume that a target ultrametric space $V$ has diameter $r>\delta$, with open root blocks $V_1,\ldots,V_k$. A distortion- $\delta$ correspondence between $U$ and $V$ partitions $U$ into nonempty pieces \[ U=U_1\sqcup\cdots\sqcup U_k \] such that each $U_i$ corresponds to $V_i$ at distortion $\delta$, and every distance between $U_i$ and $U_j$, $i\ne j$, belongs to $[r-\delta,r+\delta]$. Indeed, one source point cannot meet two target root blocks, because their mutual distance is $r>\delta$. \begin{proposition} \label{prop:local-half-barrier} Let $U,V$ be finite metric spaces with $V$ ultrametric, $\diam(U)\le\delta$, and $r=\diam(V)>\delta$. If there is a correspondence between $U$ and $V$ of distortion at most $\delta$, then \[ \delta2\delta$. \end{proof} The restriction in \cref{prop:local-half-barrier} concerns a source piece of diameter at most $\delta$. It does not impose a lower bound on the contraction ratio for arbitrary finite ultrametric spaces. \subsection{A strongly rigid perturbation} We now perturb the spaces $X_0,Y_0$ from \cref{thm:ordinary-ultrametric-contraction} so that both spaces are strongly rigid and every triangle inequality is strict. Label the source points by $0,\ldots,5$. In lexicographic edge order \[ 01,02,03,04,05,12,13,14,15,23,24,25,34,35,45, \] define the source distances to be \begin{equation} \begin{split} (&2.001,10.002,10.003,10.004,10.005, 10.006,10.007,10.008,10.009,\\ &2.010,10.011,10.012,10.013,10.014,2.015). \end{split} \label{eq:perturbed-source} \end{equation} Label the target points by $0,\ldots,4$. In edge order \[ 01,02,03,04,12,13,14,23,24,34, \] define the target distances to be \begin{equation} (10.001,10.002,10.003,10.004,10.005, 10.006,10.007,3.008,3.009,3.010). \label{eq:perturbed-target} \end{equation} \begin{theorem} \label{thm:strongly-rigid-contraction} Let $X,Y$ be the spaces defined by \eqref{eq:perturbed-source} and \eqref{eq:perturbed-target}. Then $X$ and $Y$ are strongly rigid and have only strict triangle inequalities. Moreover, \[ \dGH(X,Y)=\frac{188}{125}=1.504, \qquad \dGH(\Exp(X),\Exp(Y)) \le \frac{403}{400}=1.0075. \] \end{theorem} \begin{proof} All positive distances in either displayed list are distinct. To check the triangle inequalities for $X$, regard $\{0,1\},\{2,3\},\{4,5\}$ as its three displayed components. A triangle containing two points in one component has one edge at least $2.001$ and two cross edges whose difference is at most $0.012$. A triangle meeting all three components has all edges between $10.002$ and $10.014$. Every such triangle is strict. The verification for $Y$ is similar. The component $\{2,3,4\}$ has edge lengths $3.008,3.009,3.010$, and all cross edges lie between $10.001$ and $10.007$. Hence both spaces are strongly rigid metrics in general position. We next compute the base distance. For the sake of contradiction, suppose that a correspondence has distortion less than $3.008$, all targets related to one of the source pairs $\{0,1\},\{2,3\},\{4,5\}$ would lie in a single target component, since \[ 2.015+3.008<10.001. \] Coverage again induces a bijection between the three displayed source and target components. One source pair must cover the three-point target component $\{2,3,4\}$. Repetition of a source point then incurs at least the target separation $3.008$. Thus every base correspondence has distortion at least $3.008$. Equality is attained by the relation \[ \{(0,0),(1,0),(2,1),(3,1),(4,2),(4,3),(5,4)\}. \] The two edges with source $4$ and targets $2,3$ incur distortion $3.008$. The two collapsed pairs cost $2.001$ and $2.010$, the other within-component errors are smaller than one, and every cross-component error is at most $0.008$. Therefore the exact base distortion is $3.008$, and \eqref{eq:correspondence-formula} yields \[ \dGH(X,Y)=\frac{3.008}{2}=1.504. \] For the hyperspace upper bound, use the same seven support-block pairing as in the proof of \cref{thm:ordinary-ultrametric-contraction}. The three source blocks of cardinality $3$ collapse onto the three singleton target blocks. Their diameters are at most $2.015$. Each remaining source block has cardinality at least $9$, diameter at most $2.015$, and minimum positive distance at least $2.001$. Each remaining target block has cardinality $7$, diameter at most $3.010$, and minimum positive distance at least $3.008$. The graph of any surjection between a paired pair of these blocks has distortion at most \[ \max\{2.015,3.010-2.001\}=2.015. \] For points in distinct support blocks, the source Hausdorff distance lies in $[10.002,10.014]$, while the target Hausdorff distance lies in $[10.001,10.007]$. The cross-block error is therefore at most $0.013$. The union of the seven block correspondences has distortion at most $2.015$, and hence \[ \dGH(\Exp(X),\Exp(Y)) \le \frac{2.015}{2} =1.0075. \] This is strictly smaller than $\dGH(X,Y)$. \end{proof} The examples settle the unrestricted ordinary-distance isometry question negatively. They do not contradict hyperspace injectivity. Strict shortening of a positive distance does not identify two isometry classes, and \cref{thm:equal-cardinality-contraction} shows that unequal base cardinality is not the source of the failure. What survives for $\dGH$ is the general nonexpansion in \cref{prop:ordinary-nonexpansion}, equality on special pairs, and the open possibility of a coarse universal reverse inequality. \begingroup \sloppy \printbibliography \endgroup \end{document}